Problem: $\lim_{x\to\infty}\dfrac{(\ln(6x))^2}{(\ln(2x))^2}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $1$ (Choice C) C $9$ (Choice D) D $\infty$
Solution: $\lim_{x\to\infty} (\ln(6x))^2=\infty$ and $\lim_{x\to\infty} (\ln(2x))^2=\infty$, so $\lim_{x\to\infty}\dfrac{(\ln(6x))^2}{(\ln(2x))^2}$ results in the indeterminate form $\dfrac{\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{(\ln(6x))^2}{(\ln(2x))^2} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[(\ln(6x))^2\right]}{\dfrac{d}{dx}[(\ln(2x))^2]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{2\cdot\ln(6x)\cdot\dfrac{6}{6x}}{2\cdot\ln(2x)\cdot\dfrac{2}{2x}} \\\\ &=\lim_{x\to\infty}\dfrac{\ln(6x)}{\ln(2x)} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[\ln(6x)\right]}{\dfrac{d}{dx}[\ln(2x)]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{\left(\dfrac{6}{6x}\right)}{\left(\dfrac{2}{2x}\right)} \\\\ &=1 \end{aligned}$ Note that we used l'Hôpital's rule twice, because the first time we used it, we ended with the indeterminate form $\dfrac{\infty}{\infty}$ too. Also note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[\ln(6x)\right]}{\dfrac{d}{dx}[\ln(2x)]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{(\ln(6x))^2}{(\ln(2x))^2}=1$.